问题: Sn
Sn=(4/3)an-(1/3)*2^(n+1)+(2/3)
(1)求an,Sn
(2)Tn=2^n/Sn,求证T1+T2+T3+T4+......Tn<3/2
解答:
Sn=(4/3)an-(1/3)*2^(n+1)+(2/3)
(1)求an,Sn
(2)Tn=2^n/Sn,求证T1+T2+T3+T4+......Tn<3/2
S1=a1=(4/3)a1-(4/3)+(2/3)--->S1=a1=2
Sn = (4/3)an-(2/3)*2^n+(2/3)
S(n-1) = (4/3)a(n-1)-(1/3)*2^n+(2/3)
--->an = Sn-S(n-1) = (4/3)an - (4/3)a(n-1) - (1/3)*2^n
--->an = 4a(n-1) + 2^n
--->an/2^n = 2[a(n-1)/2^(n-1)] + 1
--->1+an/2^n = 2[1+a(n-1)/2^(n-1)]
即：数列{1+an/2^n}是等比数列，首项=1+a1/2=2，公比=2
--->1+an/2^n=2^n--->an = (2^n-1)•2^n = 4^n-2^n
--->Sn=[4^(n+1)-1]/3-[2^(n+1)-1]=(4/3)4^n-2^(n+1)+(2/3)
Tn = 2^n/Sn = 2^n/[(4/3)4^n-2^(n+1)+(2/3)]
= (3•2^n)/[4•4^n-6•2^n+2]
= (3•2^n)/[2•(2^n-1)(2•2^n-1)]
= [(3/2)•2^n•(1/2)]/[(2^n-1)(2^n-1/2)]
= [(3/2)•2^n][(2^n-1/2)-(2^n-1)]/[(2^n-1)(2^n-1/2)]
= [(3/2)•2^n][1/(2^n-1)-1/(2^n-1/2)]
= (3/2)•2^n/(2^n-1) - (3/2)2^(n+1)/[2^(n+1)-1]
--->T1+T2+T3+...+Tn
= (3/2)•2^1/(2^1-1) - (3/2)2^(n+1)/[2^(n+1)-1]
= 3 - (3/2)/[1-1/2^(n+1)]
＜3 - (3/2)
= 3/2 （证毕）
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